Question

A sample of 64 cows is used to estimate the average weight of a herd’s population. If the population standard deviation is 96 pounds, what is the margin of error of the population average weight estimate for a 90% confidence interval?

Answer 1

Solution :

Given that,

Population standard deviation =    = 96
Sample size = n =64

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * ( 96 /  64 )

=19.74